"

Putting it all together

Let’s put everything we learned above into something more concise. Closure is defined just as before. Thus:

Definition 8.4 (Closure). We call a branch of a tree closed provided there are some 1\mathcal{L}_1 formulas of form YY, ¬Y\neg Y occurring on it. We call a tree closed if all its branches are closed.

Then, we can define what syntactic entailment means in first-order logic.

Definition 8.5 (Syntactic entailment). Let A={X1,X2,...}A=\{X_1, X_2, …\} be any set of formulas and let YY be any formula of 1\mathcal{L}_1. Then, the set of premises AA syntactically entails YY, or YY is a syntactic consequence of AA (in first-order logic), iff there is a closed tree with ¬Y\neg Y at its origin, where each occurrence of a formula on the tree except ¬Y\neg Y is derived using one of the following rules:

  1. the rules ¬¬\neg\neg, \wedge, ¬\neg \vee, ¬\neg \rightarrow;

  2. the rules \vee, ¬\neg \wedge, \rightarrow;

  3. the rules \forall, \exists, ¬\neg \forall, ¬\neg \exists;

  4. the premise rule Pr (using AA).

In such cases, we write AYA \vdash Y, and call the closed tree a proof of YY from the premise set (or simply, premises) AA. If no occurrence of a formula on the tree was introduced using the rule Pr, we say YY is a theorem of first-order logic.

Soundness and completeness

Similar to zeroth-order logic, we still have that relative to the first-order semantics introduced above, the tableaux system introduced is sound and complete.

Proposition 8.8 (Soundness). For any set of formulas AA and any formula YY of 1\mathcal{L}_1, if AYA \vdash Y, then AYA \models Y.

Remark 8.2. In other words, if a formula YY is syntactically entailed by the set of formulas AA, then we can be sure that YY is also semantically entailed by AA.

Proposition 8.9 (Completeness). For any set of formulas AA and any formula YY of 1\mathcal{L}_1, if AYA \models Y, then AYA \vdash Y.

Remark 8.3. In other words, if a formula YY is semantically entailed by the set of formulas AA, then we can be sure that YY is also syntactically entailed by AA.

Now if we put these two together, we get a biconditional of the form: AY iff AYA \vdash Y \text{ iff } A \models Y Thus, syntactic and semantic entailment go hand in hand in first-order logic. If one obtains, the other obtains too!

It is easy to see how this fact entails the following:

Proposition 8.10. A⊭YA \not\models Y iff there is no tree with ¬Y\neg Y at its origin constructed with the rules enumerated above whose branches are all closed.

Proof. By soundness and completeness, we have that AYA \models Y iff AYA \vdash Y, so we also have that A⊭YA \not\models Y iff AYA \nvdash Y. Moreover, AYA \nvdash Y means that YY is not a syntactic consequence of AA. The formula YY would be a syntactic consequence of YY if there were a closed tree with ¬Y\neg Y at its origin with the rules enumerated above whose branches are all closed. So since this is not the case, there can be no such tree. ◻

License

Icon for the Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License

Learning Logic Backwards Copyright © by Peter Susanszky is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License, except where otherwise noted.